Let $h(x)=x^4-20x^3+150x^2$. For what values of $x$ does the graph of $h$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-30$ (Choice B) B $x=5$ (Choice C) C $x=15$ (Choice D) D $h$ has no points of inflection.
We can find the inflection points of the graph of $h$ by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=12(x-5)^2$. $h''(x)=0$ for $x=5$. Since $h''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection point is $x=5$. Our possible inflection points divide the number line into two intervals: $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $(-\infty,5)$ $(5,\infty)$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $(-\infty,5)$ $x=0$ $h''(0)=300>0$ $h$ is concave up $\cup$ $(5,\infty)$ $x=10$ $h''(10)=300>0$ $h$ is concave up $\cup$ We can see that the graph of $h$ does not change concavity at $x=5$. Therefore, $x=5$ is not a point of inflection. In conclusion, $h$ has no points of inflection.